Question

$\int \frac{dx}{x\sqrt{1-x^3}}=$

• A. $\frac{1}{3}\log |\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}|+c$
• B. $\frac{1}{3}\log |\frac{\sqrt{1-x^2}+1}{\sqrt{1-x^2}-1}|+c$
• C. $\frac{1}{3}\log |\frac{1}{\sqrt{1-x^3}}|+c$
• D. $\frac{1}{3}\log |1-x^3|+c$

Answer 1

The correct option is A $\frac{1}{3}\log |\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}|+c$

$\int \frac{dx}{x\sqrt{1-x^3}}=\int \frac{dx}{x^{\frac{5}{2}}\sqrt{{\left(\frac{1}{x^{\frac{3}{2}}}\right)}^2-1}}$

Put $x^{\frac{-3}{2}}=t$

$\frac{-3}{2}{x}^{\frac{-5}{2}}dx=dt⟹\frac{dx}{x^{\frac{5}{2}}}=\frac{-2}{3}dt$

$\int \frac{\frac{-2}{3}dt}{\sqrt{t^2-1}}=\frac{-2}{3}\int \frac{dt}{\sqrt{t^2-1}}$

we know, $\int \frac{1}{\sqrt{x^2-a^2}}dx=\log |x+\sqrt{x^2-a^2}|+c$

$=\frac{-2}{3}\log \mid t+\sqrt{t^2-a^2}\mid$

Put the value of $t$

$=\frac{-2}{3}\log \mid x^{\frac{-3}{2}}+\sqrt{x^{-3}-1}\mid +c$

on simplifying we get,

$=\frac{1}{3}\log \mid \frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\mid +c$